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Right Triangles

Angles in Standard Position

Sine Law & Ambiguous Case

Cosine Law

Problem Solving

Summary&Test

 

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UNIT 6 : BASIC TRIGONOMETRY WITH TRIANGLES

 LESSON 2: ANGLES IN STANDARD POSITION HOMEWORK QUESTIONS (Solutions below)

 

 

Definition: An angle is in standard position if it has its vertex at the origin and initial arm along the positive x-axis. The terminal arm is found by rotating the initial arm about the origin to a terminal position in one of the 4 quadrants. The rotation is positive if it is in the counter clockwise direction and negative if in the clockwise direction.

 

 

To find the trigonometric ratios of angles between 90o and 180o, use the following rules:

 

 

 

Text Box:
 

 

 

 

 

 


 

2. Evaluate each of the following using reference (related acute) angles. Give answers to 5 decimal places.

 

a) cos 121o b) sin 106o c) cot 97o d) sin 165.4o

 

 

3. Evaluate each of the following directly using your calculator. Give answers to 5 decimal places.

 

a) sin 1190 b) tan 143.6o c) sec 124.9o d) sin 148.2o

 

 

a) sin B = 0.3256 b) cos B = -0.87921 c) csc B = 1.5894 d) cos B = 0.3685

 

 

 

 

Solutions:

 

 

Solution:

 

 

 

x = -6

y = 8

r = 10

 

 

 

 

 

 

 

 

2. Evaluate each of the following using reference (related acute) angles. Give answers to 5 decimal places.

 

a) cos 121o b) sin 106o c) cot 97o d) sin 165.4o

 

 

Solutions: [Recall to get the reference (related acute) angle, subtract from 180o].

a) cos 121o = - cos (180o 121o)

= - cos 59o ** 59o is the reference (related acute) angle

= - 0.51504 ** Using your cacluator

 

b) sin 106o = sin (180o 106o)

= sin 74o ** 74o is the reference (related acute) angle

= 0.96126 ** Using your cacluator

 

 

b) sin 165.4o = sin (180o 165.4o)

= sin 14.6o ** 14.6o is the reference (related acute) angle

= 0.25207 ** Using your cacluator

 

 

3. Evaluate each of the following directly using your calculator. Give answers to 5 decimal places.

 

a) sin 1190 b) tan 143.6o c) cos 124.9o d) sin 148.2o

 

Solutions:

 

 

a) sin B = 0.3256 b) cos B = -0.87921 c) csc B = 1.5894 d) cos B = 0.3685

 

Solutions:

a) Since sin B is positive, < B could be in either the first or second quadrant. There are two solutions.

 

b) Since cos B is negative, < B will be in the second quadrant.

 

c) Since csc B is positive, < B could be in either the first or second quadrant. There are two solutions.

 

d) Since cos B is positive, < B will be in the first quadrant.

 

 

 

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